Let $y=\sqrt{\ln(x)}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{\dfrac{1}{x}}$ (Choice B) B $\dfrac{\left(\dfrac{1}{2\sqrt x}\right)}{\sqrt x}$ (Choice C) C $\dfrac{1}{\sqrt x}$ (Choice D) D $\dfrac{1}{2x\sqrt{\ln(x)}}$
Explanation: $\sqrt{\ln(x)}$ is a composition of two, more basic, functions: $\ln(x)$ and $\sqrt x$. In other words, suppose $u(x)=\ln(x)$ and $v(x)=\sqrt x$, then $\sqrt{\ln(x)}=v\bigl(u(x)\bigr)$, or $(v\circ u)(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=\sqrt x$, and therefore $v'(x)=\dfrac{1}{2\sqrt x}$. Now we plug $u(x)=\ln(x)$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\Bigl(\ln(x)\Bigr) \\\\ &={\dfrac{1}{2\sqrt{\ln(x)}}} \end{aligned}$ Finding $u'(x)$ $u(x)=\ln(x)$, and therefore $u'(x)={\dfrac{1}{x}}$. Putting things together $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\sqrt{\ln(x)}\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=\ln(x)\text{, }v(x)=\sqrt x} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={\dfrac{1}{2\sqrt{\ln(x)}}}\cdot {\dfrac{1}{x}} \\\\ &=\dfrac{1}{2x\sqrt{\ln(x)}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{1}{2x\sqrt{\ln(x)}}$.